Re: Kernel stack

[Jan Hudec]

On Tue, Oct 12, 2004 at 19:05:52 +0900, aq wrote:
> > There is no SS entry, because SS does not specify the stack. It is siply
> > a segment in which the stack lives. Any segment, that covers all address
> > space will do! IIRC in kernel SS == DS.
> yes, if I am right, in Linux SS, DS and CS all point to the same base
> address ( = 0 ?). To be exact, SS !=DS, since the segment registers in
> protected mode point to segment selectors (in GDT ?), and we should
> compare the value stored GDT entries, not the value of SS and DS in
> this case.

Of course. Though since DS and SS need the same parameters, they might
actualy point to the same GDT entry. I don't know if they actualy do,
though.

By the way, C compilers usualy set SS and DS with same base. They would
have to do conversion when taking pointers to local variables otherwise.

> > The kernel stack is allocated together with the task_struct. Two pages
> > are allocated and task_struct is placed at the start while the stack is
> > placed at the end and grows down towards the task_struct.
> 2 pages of kernel stack or not is optional. Recently version of kernel
> allow you to choose to use 4K or 8K size for kernel stack.

Yes, it does. Few people touch the "Kernel Hacking" though...

> >From what you all discuss, I can say: kernel memory is devided into 2
> part, and the upper part are shared between processes. The below part
> (the kernel stack, or 8K traditionally) is specifict for each process.
> 
> Is that right?

No, it's not. There is just one kernel memory. In it each process has
it's own task_struct + kernel stack (by default 8K). There is no special
address mapping for these, nor are they allocated from a special area.

When a context of some process is entered, esp is pointed to the top of
it's stack. That's exactly all it takes to exchange stacks.

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						 Jan 'Bulb' Hudec <bulb@xxxxxx>

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